Ich habe immer das folgende Skript zum Hochladen von Dateien in klassischem ASP verwendet, aber es hat nicht mehr funktioniert.

vbscript Laufzeitfehler 800a01a8
Objekt erforderlich 'Item (...)'

Ich habe ein wenig nachgeforscht und denke, dass das Problem in der Datei upload.asp mit der Funktion BuildUploadRequest liegt, aber ich kann wirklich nicht verstehen, warum

bilde

<form method="POST" action="landing-page.asp" ENCTYPE="multipart/form-data">
    <input type="file" name="file">
    <input type="hidden" name="key" value="0">
    <input type="submit" name="send" value="1">
</form>

Seite wo das Formular landet

byteCount = Request.TotalBytes
RequestBin = Request.BinaryRead(byteCount)

Dim UploadRequest
Set UploadRequest = CreateObject("Scripting.Dictionary")
BuildUploadRequest(RequestBin)  '//function defined in upload.asp
if UploadRequest.Item("key").Item("Value")="0" then  '//this is the line giving the error
    '//code here...
end if

upload.asp

Sub BuildUploadRequest(RequestBin)
    PosBeg = 1  
    PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
    boundary = MidB(RequestBin,PosBeg,PosEnd-PosBeg)    
    boundaryPos = InstrB(1,RequestBin,boundary)

    '//Get all data inside the boundaries
    Do until (boundaryPos=InstrB(RequestBin,boundary & getByteString("--")))
        '//Members variable of objects are put in a dictionary object
        Dim UploadControl
        Set UploadControl = CreateObject("Scripting.Dictionary")
        '//Get an object name
        Pos = InstrB(BoundaryPos,RequestBin,getByteString("Content-Disposition"))
        Pos = InstrB(Pos,RequestBin,getByteString("name="))
        PosBeg = Pos+6
        PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(34)))
        Name = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
        PosFile = InstrB(BoundaryPos,RequestBin,getByteString("filename="))
        PosBound = InstrB(PosEnd,RequestBin,boundary)
        '//Test if object is of file type
        If  PosFile<>0 AND (PosFile<PosBound) Then
            '//Get Filename, content-type and content of file
            PosBeg = PosFile + 10
            PosEnd =  InstrB(PosBeg,RequestBin,getByteString(chr(34)))
            FileName = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
            '//Add filename to dictionary object
            UploadControl.Add "FileName", FileName
            Pos = InstrB(PosEnd,RequestBin,getByteString("Content-Type:"))
            PosBeg = Pos+14
            PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
            '//Add content-type to dictionary object
            ContentType = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
            UploadControl.Add "ContentType",ContentType
            '//Get content of object
            PosBeg = PosEnd+4
            PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
            Value = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
        Else
            '//Get content of object
            Pos = InstrB(Pos,RequestBin,getByteString(chr(13)))
            PosBeg = Pos+4
            PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
            Value = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
        End If
        '//Add content to dictionary object
        UploadControl.Add "Value" , Value   
        '//Add dictionary object to main dictionary
        '//response.write name & "<br>"
        UploadRequest.Add name, UploadControl   
        '//Loop to next object
        BoundaryPos=InstrB(BoundaryPos+LenB(boundary),RequestBin,boundary)
    Loop
End Sub

'//String to byte string conversion
Function getByteString(StringStr)
  For i = 1 to Len(StringStr)
    charx = Mid(StringStr,i,1)
    getByteString = getByteString & chrB(AscB(charx))
  Next
End Function

'//Byte string to string conversion
Function getString(StringBin)
 getString =""
 For intCount = 1 to LenB(StringBin)
    getString = getString & chr(AscB(MidB(StringBin,intCount,1))) 
 Next
End Function

Dieser Code hat in jedem Projekt immer richtig funktioniert, aber jetzt funktioniert er nicht überall. Ich kann also nicht einfach eine andere Funktion bearbeiten und verwenden. Ich muss verstehen, warum sie nicht mehr funktioniert.